How do you simplify ${cos}^{4} α + 2 {cos}^{2} α {sin}^{2} α + {sin}^{4} α$ $cos^4alpha+2cos^2alphasin^2alpha+sin^4alpha$ ?

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Answer 1 · 2016-12-08T14:15:43

1

consider

${(x + y)}^{2} = x^{2} + 2 x y + y^{2}$ $(x+y)^2=x^2+2xy+y^2$

let $x = {cos}^{2} α y = {sin}^{2} α$ $" " x=cos^2alpha" "y=sin^2alpha " "$ on RHS

${cos}^{4} α + 2 {cos}^{2} α {sin}^{2} α + {sin}^{4} α$ $cos^4alpha+2cos^2alphasin^2alpha+sin^4alpha$

we have
$= {({cos}^{2} α + {sin}^{2} α)}^{2} = 1^{2} = 1$ $=(cos^2alpha+sin^2alpha)^2=1^2=1$